1. Sec:5.2
The Bisection Method

2. Sec:5.2 The Bisection Method
The root-finding problem is a process involves finding a root, or solution, of an equation of the form
𝑓 𝑥 = 0
for a given function 𝑓 . A root of this equation is also called a zero of the function 𝑓 .
In graph, the root (or zero) of a function is the x-intercept
Two numerical methods for
root-finding
Sec(5.2): The Bisection Method
root
Sec(6.2): The Newton-Raphson
Method

3. Sec:5.2 The Bisection Method
This technique is based on the Intermediate Value Theorem
Example:
Suppose 𝑓 is a continuous function defined on the interval [𝑎, 𝑏], with 𝑓 (𝑎) and 𝑓 (𝑏) of opposite sign. The Intermediate Value Theorem implies that a number 𝑝 exists in (𝑎, 𝑏) with 𝑓 ( 𝑝) = 0.
Show that
𝑓 (𝑥) = 10 𝑥 6 −149 𝑥 5 +10𝑥−149 10( 𝑥 4 +1)
has a root in [12, 16]
Sol: 12 16
𝒇(𝟏𝟐)=−𝟑𝟒.𝟖
𝒇(𝟏𝟔)=𝟏𝟕.𝟔

4. Sec:5.2 The Bisection Method
Example:
Use Bisection method to find the root of the function
𝑓 (𝑥) = 10 𝑥 6 −149 𝑥 5 +10𝑥−149 10( 𝑥 4 +1)
in [12, 16] 12 16
Change of sign
-34.8
17.6
𝒙 𝟏 =
True root: 𝑥 𝑟 ∗ =14.9 12 14
Change of sign 16
Iter1 𝒏
𝒙 𝒏
-34.8
-12.6
17.6
𝒙 𝟐 = 14 15 16
Iter2
Change of sign
-12.6
1.5
17.6
𝒙 𝟑 =
Change of sign
14.5 14 15
Iter3
-5.8
1.5
-12.6

5. Sec:5.2 The Bisection Method
Textbook notations
𝒙 𝒍 𝟎
𝒙 𝒖 𝟎 12 16
Change of sign
∆ 𝒙 𝟎 = 𝒙 𝒖 𝟎 − 𝒙 𝒍 𝟎
At the n-th iteration:
-34.8
17.6
endpoints of the inteval
𝒙 𝒍 𝟏
𝒙 𝒖 𝟏 12 14
Change of sign 16
[ 𝒙 𝒍 𝒏 , 𝒙 𝒖 𝒏 ]
Iter1
∆ 𝒙 𝟏 = 𝒙 𝒖 𝟏 − 𝒙 𝒍 𝟏
-34.8
-12.6
17.6
Length of the interval
𝒙 𝒍 𝟐
𝒙 𝒖 𝟐 14 15 16
∆ 𝒙 𝒏 = 𝒙 𝒖 𝒏 − 𝒙 𝒍 𝒏
Iter2
Change of sign
∆ 𝒙 𝟐 = 𝒙 𝒖 𝟐 − 𝒙 𝒍 𝟐
-12.6
1.5
17.6
= ∆ 𝒙 𝟎 𝟐 𝒏
𝒙 𝒍 𝟑
𝒙 𝒖 𝟑
Change of sign
14.5 14 15
Iter3
∆ 𝒙 𝟑 = 𝒙 𝒖 𝟑 − 𝒙 𝒍 𝟑
-5.8
1.5
-12.6

6. Sec:5.2 The Bisection Method
Error Estimates for Bisection 12 16
Change of sign
-34.8
17.6
𝒙 𝒓 ∗
At the iter1:
𝒙 𝟏 =
True root live inside this interval
𝒙 𝟏 − 𝒙 𝒓 ∗ < length of the interval 12 14
Change of sign 16
𝒙 𝟏 − 𝒙 𝒓 ∗ <∆ 𝒙 𝟏 = ∆ 𝒙 𝟎 𝟐
Iter1
-34.8
-12.6
17.6
𝒙 𝟐
At the iter2: = 14 15 16
𝒙 𝟐 − 𝒙 𝒓 ∗ < length of the interval
Iter2
Change of sign
𝒙 𝟐 − 𝒙 𝒓 ∗ <∆ 𝒙 𝟐 = ∆ 𝒙 𝟎 𝟐 𝟐
-12.6
1.5
17.6
At the nth iteration:
𝒙 𝒏 − 𝒙 𝒓 ∗ < 𝒃−𝒂 𝟐 𝒏
Error Estimates for Bisection
𝒙 𝒏 − 𝒙 𝒓 ∗ < length of the interval
𝑬 𝒂 𝒏 < 𝒃−𝒂 𝟐 𝒏
the absolute error in the n-th iteration
< 𝒃−𝒂 𝟐 𝒏
𝒙 𝒏 − 𝒙 𝒓 ∗ <∆ 𝒙 𝒏 = ∆ 𝒙 𝟎 𝟐 𝒏
= 𝒃−𝒂 𝟐 𝒏

7. Sec:5.2 The Bisection Method
Error Estimates for Bisection
𝑬 𝒂 𝒏 < 𝒃−𝒂 𝟐 𝒏 𝒏
𝒙 𝒏
𝑬 𝒂 𝒏
e-01
e-01
e-01
e-01
e-02
e-02
e-03
e-03
e-03
e-03
e-04
e-04
e-05
e-04
e-05
e-05
If 𝐸𝑎,𝑑 =10 −4 is the desired error, this equation can be solved for 𝑛
𝑬 𝒂 𝒏 < 𝟏𝟔−𝟏𝟐 𝟐 𝒏 < 𝟏𝟎 −𝟒
𝟐 𝒏 > 𝟒 𝟏𝟎 −𝟒
𝟐 𝒏 >𝟒× 𝟏𝟎 𝟒
𝒏> 𝒍𝒐𝒈 𝟒× 𝟏𝟎 𝟒 𝒍𝒐𝒈 𝟐 =𝟏𝟓.𝟐

8. Stopping Criteria Sec:5.2 The Bisection Method
function [xr,err,yc,iter,x]=bisect_ver1(f,a,b,es)
%Input: f is the function, a, b are endpts
% es is the tolerance, imax is max iter
%Output: c is the zero, yc= f(c)
ya=f(a); yb=f(b); iter =0;
if ya*yb > 0,return,end
for k=1:1000
iter = iter +1;
xr=(a+b)/2; yc=f(xr); x(k)=xr;
if yc==0
a=xr; b=xr;
elseif yb*yc>0
b=xr; yb=yc;
else
a=xr; ya=yc;
end
if b-a < es, break,end
xr=(a+b)/2; err=abs(b-a); yc=f(xr);
function [xr,err,yc,iter,x]=bisect_ver2(f,a,b,es)
%Input: f is the function, a, b are endpts
% es is the tolerance, imax is max iter
%Output: c is the zero, yc= f(c)
ya=f(a); yb=f(b); iter =0;
if ya*yb > 0,return,end
max1=1+round((log(b-a)-log(es))/log(2));
for k=1:max1
iter = iter +1;
xr=(a+b)/2; yc=f(xr); x(k)=xr;
if yc==0
a=xr; b=xr;
elseif yb*yc>0
b=xr; yb=yc;
else
a=xr; ya=yc;
end
% if b-a < es, iter=k; break,end
xr=(a+b)/2; err=abs(b-a); yc=f(xr);
a=12; b=16; es=1e-4;
% [xr,err,yc,iter,x]=bisect_ver2(f,a,b,es);
[xr,err,yc,iter,x]=bisect_ver1(f,a,b,es);
iteration = [1:iter]';
res = [ iteration, x' , x'-14.9]
fprintf(' %d %14.10f %14.10e \n', res');

9. Sec:5.2 The Bisection Method